of 1 34
The Elements As Mathematical Constructs
Essays where anthropological cosmology
meet with cosmology in physics
Genesis Project California 2021!
⋅
⋅
of 2 34
Abstract
If the elements are mathematical constructs, then we need to as well explain mass as a
mathematical construct. Both ventures are taken on here.
of 3 34
Important
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the elements. As
you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and germanium are in
group 14 meaning they have 4 valence electrons and want 4 for more to attain noble gas electron
configuration. If we dope Si with B from group 13 it gets three of the four electrons and thus has a
deficiency becoming positive type silicon and thus conducts. If we dope the Si with P from group 15 it
has an extra electron and thus conducts as well. If we join the two types of silicon we have a
semiconductor for making diodes and transistors from which we can make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the right of Si
but doping agent B is not directly to the left, aluminum Al is. This becomes important. I call (As-Ga) the
differential across Ge, and (P-Al) the differential across Si and call Al a dummy in the differential because
boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with subscripts
that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the first row
and has 4 valence electrons making it carbon (C). I believe that the AI elements can be organized in a 3 by
3 matrix makes them pivotal to structure in the Universe because we live in three dimensional space so
the mechanics of the realm we experience are described by such a matrix, for example the cross product.
Hence this paper where I show AI and biological life are mathematical constructs and described in terms
of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are carbon and
nitrogen respectively, there is every reason to proceed with this paper if the idea is to show not only are
the AI elements and biological elements mathematical constructs, they are described in terms of one
another. We see this because the first row is ( B, C, N) and these happen to be the only elements that are
not core AI elements in the matrix, except boron (B) which is out of place, and aluminum (Al) as we will
see if a dummy representative makes for a mathematical construct, the harmonic mean. Which means we
have proved our case because the first row if we take the cross product between the second and third rows
are, its respective unit vectors for the components meaning they describe them!
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 4 34
The Computation
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological elements C
and N being the unit vectors are multiplied by the AI elements, meaning they describe them! But we have
to ask; Why does the first row have boron in it which is not a core biological element, but is a core AI
element? The answer is that boron is the one AI element that is out of place, that is, aluminum is in its
place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
Equation 1.
A = (Al, Si, P )
B = (G a, G e, As)
A ×
B =
B
C
N
Al Si P
G a Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g /m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g /m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsi nθ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si( A s − G a) + G e(P − Al )
SiG e
=
2B
Ge + Si
of 5 34
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy (note: using
an older value for Ge here, it is now 72.64 but that makes the equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 2.
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Equation 3.
Equation 4.
If in the equation (The accurate harmonic mean form):
Equation 5.
We make the approximation
Equation 6.
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d xd y ≈
1
Ge − Si
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al )d xdz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
Si
B
(As − G a) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
of 6 34
Then the Stokes form of the equation becomes
Equation 7.
Thus we see for this approximation there are two integrals as well:
Equation 8.
Equation 9.
For which the respective paths are
One of the double integrals on the left is evaluated in moles per grams, the other grams per mole
(0 to 1 moles per gram and 0 to 1 grams per mole).
The Geometric Interpretation…
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d yd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d yd z =
2
3
∫
Ge
Si
dz
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
of 7 34
of 8 34
By making the approximation
In
We have
Equation 10.
is the differential across Si, is the differential across
Ge and is the vertical differential.#
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor elements by molar mass which are used to make
circuits.
We say (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 11.
2SiGe
Si + Ge
≈ Ge − Si
Si(As − G a)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ G e
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
of 9 34
Thus since
And we have
Equation 12.
We see and are both and c is in the Si (silicon) field wave, but for E and B fields c is the speed of
light.
To find the Si wave our differentials are
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔGe =
ΔS
Si
B −
Ge
Si
ΔSi
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
Ge
Si
= μϵ
0
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Ge = 0
of 10 34
It is amazing how accurately we can fit these differentials with and exponential equation for the upward
increase. The equation is
This is the halfwave:
ΔC = N − B = 14.01 − 10.81 = 3.2
Δ Si = P − Al = 30.97 − 26.98 = 3.99
ΔG e = A s − G a = 74.92 − 69.72 = 5.2
Δ Sn = Bi − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 11 34
Equation 13.
Interestingly, the 0.4 is boron (B) over aluminum (Al) the very two elements that lead us to looking for a
wave equation because boron was the out of place element in the AI periodic table that lead to us using
aluminum as its dummy representative in the Si differential and that itself divided into the left hand terms
to give us the harmonic mean between the central AI elements semiconductor materials Si and Ge. The
Ag and Cu are the central malleable, ductile, and conductive metals used in making electrical wires to
carry a current in AI circuitry.
Building Another Matrix #
We pull these AI elements out of the periodic table of the elements to make an AI periodic
table:#
We now notice we can make a 3 by 3 matrix of it, which lends itself to to the curl of a vector
field by including biological elements carbon C (above Si):#
=#
=#
#
Which resulted in Stokes theorem (Beardsley, Essays In Cosmic Archaeology Volume 3):#
y(x) = e
B
Al
x
+
Ag
Cu
B
Al
=
10.81
26.98
= 0.400667
Ag
Cu
=
107.87
63.55
= 1.6974 ≈ 1.7
i
j
k
∂
∂x
∂
∂y
∂
∂z
(−C ⋅ P)y (Si ⋅ Ga)z (Ge ⋅ As)y
(Ge ⋅ As − Si ⋅ G a)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
of 12 34
Equation 14. #
Where#
#
#
#
We were then able to write this with product notation#
Equation 15. #
While we have the AI BioMatrix#
Which we used to formulate a similar equation (Beardsley, Essays In Cosmic Archaeology
Volume 2)#
We can form another 3X3 matrix we will call the electronics matrix (Beardsley, Cosmic
Archaeology, Volume Three):#
#
We can remove the 5th root sign in the above equation by noticing#
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = exp
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
∇ ×
u = (Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
d
a =
(
zdyd z
i + yd yd z
k
)
u = (−C ⋅ P)y
i + (Si ⋅ Ge)z
j + (G a ⋅ As)y
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
n
n
∏
i=1
x
i
of 13 34
Equation 16. #
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=#
#
Where we have substituted carbon (C=12.01) the core biological element for copper (Cu).#
But since we have:#
Equation 17. #
We take the ratio and have#
#
Almost exactly 3 which is the ratio of the perimeter of regular hexagon to its diameter used to
estimate pi in ancient times by inscribing it in a circle:#
#
Perimeter=6#
Diameter=2#
6/2=3#
#
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ Cu ⋅ Ag ⋅ Au
523,818,646.5
g
5
mol
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = 170,535,359.662(g/mol )
5
523,818,646.5
170,535,359.662
= 3.0716
π = 3.141...
of 14 34
Thus we have the following equation…#
Equation 18. #
Showing The Calculation using the most accurate data possible…#
Ge=72.64#
As=74.9216#
Si=28.085#
Ga=69.723#
C=12.011#
P=30.97376200#
=#
=#
#
#
#
#
#
=154,082,837.980+16,452,521.6822=#
#
=#
(28.085)(72.64)(12.085)(107.8682)(196.9657)=#
π
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
5
∏
i=1
x
i
(Ge ⋅ As − Si ⋅ G a)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
)
⋅
(
zdyd z
i + yd yd z
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
⋅ zdzd y + 372.025855
(
g
mol
)
2
⋅ yd zd y
)
∫
Ge
Si
3,484.134569
(
(72.64 − 28.085)
2
2
)
dy +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
mol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
mol
)
3
(
(72.64 − 28.085)
2
2
)
170,535,359.662
(
g
mol
)
5
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
of 15 34
#
Where we have substituted carbon C=12.01 for copper Cu. We use Cu, Ag, Au because they
are the middle column of our electronics matrix, they are the finest conductors used for
electrical wire. We use C, Si, Ge because they are the middle column of our AI Biomatrix. Si
and Ge are the primary semiconductor elements used in transistor technology (Artificial
Intelligence) and C is the core element of biological life. We have#
#
#
Perimeter/Diameter of regular hexagon = 3.00#
#
The same value as our 3.0716 if taken at two places after the decimal.!
523,818,646.5
g
5
mol
5
523,818,646.5
170,535,359.662
= 3.0716
π = 3.141...
3.141 + 3.00
2
= 3.0705
of 16 34
!
Plot of the Surface Area
of 17 34
Atomic Number A very convenient way to estimate pi is with the regular hexagon because its
side is the same as its radius. Thus if its side is one then its radius is one meaning its perimeter
is six and its diameter is two giving the integer three even:#
#
We see this can very accurately be approximated by averaging the regular
pentagon with the regular octagon.#
#
The sum of the angles is . .
, , and 54+36=90 where
. a=0.68819096. We have and
then the diameter D is .#
The perimeter P over the diameter is . By similar reasoning we
have for a regular octagon:#
#
. The angle #
Thus,…#
#
Is a regular hexagon.#
We see that the atomic radio of silicon the core element of artificial intelligence (transistor
technology) fits together with the core element of biological life carbon if the silicon is taken as
inscribed in a regular dodecagon and the carbon is taken as inscribed in a regular octagon. We
have:#
, , #
, , #
A = 180
∘
(n − 2)
A = 180
∘
(3) = 540
∘
540
∘
5
= 108
∘
108
∘
2
= 54
∘
a /s
2
= ta n54
∘
2 cos36
∘
= Φ
a
2
+ (s /2)
2
= 0.850650808
D = 1.701301617 ≈ 3
P/D = 2.938926261 ≈ 3
P/D = 3.061457459 ≈ π
22.5
∘
= 0.41421 ≈ 2 − 1
3.061407459 + 2.93892621
2
= 3.00019686 ≈ = 3.00000 = 3
D = 1 + 2x
x
2
+ x
2
= 1
2
2x
2
= 1
2x
2
= 1
x = 2/2
D = 1 + 2
of 18 34
Apothem: For a regular dodecagon:#
=#
The radius of a silicon atom is Si=0.118nm and that of carbon is
C=0.077nm:#
#
#
#
This has an accuracy %#
It makes sense that we define molar mass in terms of carbon because being element six it can
be thought of in terms of our regular hexagon which defines pi as an integer, the integer 3 (that
is to say as a whole number, 3 is not a fraction in that there is nothing but zeros after the
decimal). Carbon can be thought of in terms of the regular hexagon because it describes the
closest packing of equal radius spheres (a so-called “six-around-one”):#
But though carbon may be six protons, it has six neutrons giving it a molar mass of 12.01
approximately twice it atomic number of six. But twelve is our dodecagon which inscribes
silicon if it is to line up with carbon as the regular octagon. And here it is interesting to note that
carbon is made in stars by combining beryllium-8 with helium, the eight of our regular octagon.#
Theory For Inertia
In order to present the elements as mathematical structures we need to explain the matter from
which they are made as mathematical constructs. We need a theory for Inertia. I had found
(Beardsley Essays In Cosmic Archaeology. 2021) where I suggested the idea of proton
seconds, that is six protons-seconds, which is carbon the core element of biological life if we
can figure out a reason to divide out the seconds. I found#
Equation 19. #
Where h is Planck’s constant 6.62607E—34 Js, r_p is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
a = (1 + 2)/2 = 1.2071
a =
s/2
tan(θ /2)
=
0.5
tan15
∘
(15
∘
)
Si
C
=
0.118
0.077
= 1.532
a
S
i
a
C
=
1.866
1.2071
= 1.54585
1.532
1.54585
100 = 99
1
t
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
of 19 34
299,792,459 m/s. And t_1 is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is .#
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: #
#
Matter is that which has inertia. This means it resists change in position with a force
applied to it. The more of it, the more it resists a force. We understand this from
experience, but what is matter that it has inertia?
I would like to answer this by considering matter in one of its simplest manifestations,
the proton, a small sphere with a mass of 1.6726E-27 kg. This is a measure of its inertia.
I would like to suggest that matter, often a collection of these protons, is the three
dimensional cross-section of a four dimensional hypersphere.
The way to visualize this is to take space as a two-dimensional plane and the proton as a
two dimensional cross-section of a sphere, which would be a circle.
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a
two dimensional plane. As such there has to be a normal vector holding the higher
dimensional sphere in a lower dimensional space. Thus if we apply a force to to the cross-
section of the sphere in the plane there should be a force countering it proportional to
the normal holding it in a lower dimensional universe. This counter force would be
experienced as inertia. It may even induce in it an electric field, and we can see how it
may do the same equal but opposite for the electron. Refer to the illustration on the
following page…
α
m
p
= 1.67262E − 27kg
α
2
=
U
e
m
e
c
2
of 20 34
of 21 34
The interesting thing is, having seen this connected to meteors I wrote a program for
that study, and what results in the first part of it is quite interesting. Here is the output:
How many values would you like to calculate for t in equation 1 (no
more than 10?): 8
Enter a number for t 1
: 1
protons 6.0300 protons
Enter a number for t 2
: 2
protons 3.0150 protons
Enter a number for t 3
: 3
protons 2.0100 protons
Enter a number for t 4
: 4
protons 1.5075 protons
Enter a number for t 5
: 5
protons 1.2060 protons
Enter a number for t 6
: 6
protons 1.0050 protons
Enter a number for t 7
: 7
protons 0.8614 protons
Enter a number for t 8
: 8
protons 0.7537 protons
The interesting thing we see is that t equal to 1 second yields six
protons, is the number of protons in carbon, which is the element upon
which life as we know it is based and if t is equal to six seconds,
then the result is 1.0050 rounds to 1.01 is the molar mass of hydrogen
and the number of protons in hydrogen which is one and hydrogen, is
the first element in the periodic table, the most abundant element in
the Universe, and combines with carbon to make hydrocarbons the basic
skeleton of organic matter. So we see we have the following
fascinating result:
(6 protons)(1 second)=(6 seconds)(1 proton)
1
α
2
m
p
h 4π r
2
p
G c
= (6pr ot on s)(1secon d )
1
α
2
m
p
h 4π r
2
p
G c
= (1pr ot on)(6seco n d s)
of 22 34
Here is the section of code I wrote in C so you can verify the
results:
//
// main.c
// Asteroids
//
// Created by Ian Beardsley on 11/5/21.
// Copyright © 2021 Ian Beardsley. All rights reserved.
//
#include <stdio.h>
#include <math.h>
int i, n;
float t,p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
int main(int argc, const char * argv[])
{{
do
{
printf("How many values would you like to calculate for t in
equation 1 (no more than 10?): ");
scanf("%i", &n);
}
while (n>=10);
for(i=1; i<=n; i++)
{
printf("Enter a number for t %i\n: ", i);
scanf("%f", &t);
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/
(G*c));
printf("protons %.4f protons \n", protons[i]);
}
of 23 34
Showing the units are correct for equation 1, and the
numerical computation by hand for it.
of 24 34
Note: #
Concerning the two equations#
#
#
While I understand that one second is a human invention and can’t be taken as significant, the
equations have meaning in that there are two equations each utilizing a second so they are
connected. But what really makes me wonder is how one can predict carbon, the other
hydrogen so accurately with the unit of a second. It was a conspiracy on the part of those who
formulated the duration of a second a long time ago to be what it is? The Ancient Egyptians,
The Babylonians, The Julian calendar—who, what, when?#
6 protons gives a little more than a second. This makes a shorter day. We have#
Equation 20. #
H=1.00784 g/mol, carbon = 6 protons#
h=6.62607E-34, r_p=0.833E-15, G=6.67408E-11, c=299,792,459#
(24hours)/(1.004996352)=23.8806837hours/day#
Over the past several billion years the length of the year has not deviated much from 365.25
days because of Kepler’s period for the orbit of a planet but, because the earth loses energy to
the moon, its days becomes longer over time by 0.0067 hours per million years which, we can
see from examining sedimentation band growth, which follows the lunar month. To get our
23.8806837 hours per day we have to go back about 18 million years:#
24-23.8806837=0.0067t#
t=17.80840299 million years#
This is the Miocene, a time which the Earth cooled slowly towards a series of ice ages,
between the Oligocene and the Pliocene defined by the boundaries of the cooler Oligocene
and the warmer Pliocene. See Messinian Salinity Crisis and Zarclean Flood. When the
dinosaurs roamed the Earth 70 million years ago the day was 23.7 hours long, about a half an
hour shorter. They seemed to have gone extinct about 65 million years ago when an asteroid
hit in the Yucatan, probably because we were moving through a place in the galaxy full of
debris, through which we pass every 27 million years. But asteroids in the asteroid belt pose a
threat as well, and that is why we catalog their trajectories. The asteroid belt is about 1AU thick
at 2.2 to 3.2 AU from the sun, the earth-sun separation being 1 AU. Carbon-12 the basis of life
as we know it is 12.01 g/mol. It is made in stars from Beryllium-8 is 8.0053051 g/mol. Thus we
have#
1
α
2
m
p
h4π r
2
p
Gc
= (6protons)(1second )
1
α
2
m
p
h4π r
2
p
Gc
= (1proton)(6secon ds)
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
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#
#
Delta x equal to 1 AU is both Earth and The Asteroids. Mars is at 1.52AU. Delta X cancels with
the Earth leaving Mars equal to carbon to beryllium, which is life, Does this say we need to be
able to colonize Mars to succeed as a species?#
#
Or, does it mean we need to put bases on the moon to mine Helium-3 as a clean, renewable
energy source.#
Using my Asteroids 01 program
//
// main.c
// Asteroids 01
//
// Created by Ian Beardsley on 11/12/21.
// Copyright © 2021 Ian Beardsley. All rights reserved.
//
#include <stdio.h>
#include<math.h>
int main(int argc, const char * argv[]) {
int i, n;
float t,p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("How many values would you like to calculate for t in
equation 1 (no more than 10?): ");
scanf("%i", &n);
}
while (n>=100);
for(i=1; i<=n; i++)
{
printf("Enter a number for t %i\n: ", i);
scanf("%f", &t);
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/
(G*c));
printf("protons %.4f protons \n", protons[i]);
}}
12
C
8
Be
=
Mars
Ear th
Δx
Δx = 1AU
12
C =
4
He +
8
Be
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We can play with our equation
To see the output…
Asteroids 01 output#
Enter a number for t 1
: 0.25
protons 24.1199 protons
Enter a number for t 2
: 0.50
protons 12.0600 protons
Enter a number for t 3
: 0.618
protons 9.7572 protons
Enter a number for t 4
: 0.666666667
protons 9.0450 protons
Enter a number for t 5
: 0.75
protons 8.0400 protons
Enter a number for t 6
: 1.0
protons 6.0300 protons
Enter a number for t 7
: 1.25
protons 4.8240 protons
Enter a number for t 8
: 1.414
protons 4.2645 protons
Enter a number for t 9
: 1.5
protons 4.0200 protons
Enter a number for t 10
: 1.6
protons 3.7687 protons
Enter a number for t 11
: 1.618
protons 3.7268 protons
Enter a number for t 12
: 1.75
protons 3.4457 protons
Enter a number for t 13
: 2.0
1
t
1
α
2
m
p
h 4π r
2
p
G c
= 6pr oton s
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protons 3.0150 protons
Enter a number for t 14
: 2.25
protons 2.6800 protons
Enter a number for t 15
: 2.5
protons 2.4120 protons
Enter a number for t 16
: 2.75
protons 2.1927 protons
Enter a number for t 17
: 3.00
protons 2.0100 protons
Enter a number for t 18
: 3.25
protons 1.8554 protons
Enter a number for t 19
: 3.5
protons 1.7229 protons
Enter a number for t 20
: 3.75
protons 1.6080 protons
Enter a number for t 21
: 4.0
protons 1.5075 protons
Enter a number for t 22
: 4.25
protons 1.4188 protons
Enter a number for t 23
: 4.5
protons 1.3400 protons
Enter a number for t 24
: 4.75
protons 1.2695 protons
Enter a number for t 25
: 5.0
protons 1.2060 protons
How many values would you like to calculate for t in equation 1 (no
more than 10?): 5
Enter a number for t 1
: 6.0
protons 1.0050 protons
Enter a number for t 2
: 6.25
protons 0.9648 protons
Enter a number for t 3
: 6.50
protons 0.9277 protons
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Enter a number for t 4
: 7
protons 0.8614 protons
Enter a number for t 5
: 8
protons 0.7537 protons
We see because the duration of the second was so wisely defined that
we can create a convenient system of units that makes time inversely
proportional to protons by a factor of 6 that we can call the proton-
seconds system of units. It predicts several key elements that can be
organized into many different matrices on which we can do operations
in terms of molar mass, density, and atomic radius. Here is the graph
I made…
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!
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We are interested after running our program a couple times those
results which are close to whole numbers of protons for values of time
t, equal to multiples of 0.25=1/4, 0.05=1/2, 0.666667=2/3,and 1.0.
which is exactly where the whole numbers seem to appear.
Enter a number for t 1
: 0.25
protons 24.1199 protons
Chromium (Cr)
Enter a number for t 2
: 0.50
protons 12.0600 protons
Magnesium (Mg)
Enter a number for t 3
: 0.666666667
protons 9.0450 protons
Flourine (F)
Enter a number for t 4
: 0.75
protons 8.0400 protons
Oxygen (O)
Enter a number for t 5
: 1.0
protons 6.0300 protons
Carbon (6)
Enter a number for t 6
: 1.5
protons 4.0200 protons
Beryllium (Be)
Enter a number for t 7
: 2.0
protons 3.0150 protons
Lithium (Li)
Enter a number for t 8
: 3.00
protons 2.0100 protons
Helium (He)
Enter a number for t 9
: 6.0
protons 1.0050 protons
Hydrogen (H)
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This is the following 3 by 3 matrix:
Equation 21.
=
And,…
In conclusion if this is supposed to be pi, and it is very close to
it, then we have measured the molar masses of these elements in the
laboratory very accurately.
Not does this only mean the second is related to the earth orbital period, but now to other
things; By dividing the day into 24 hours, the hour into 60 minutes, and the minute into 60
seconds, the second is 1/86400 of day. By doing this we have a twelve-hour daytime at spring
and fall equinox on the equator, 12 being the most divisible number for its size (smallest
abundant number). That is to say that twelve is evenly divisible by 1,2,3,4,6 which precede it
and 1+2+3+4+6=16 is greater than twelve. As such there is about one moon per 30 days and
about 12 moons per year (per each orbit) giving us a twelve-month calendar. This is all further
convenient in that the moon and earth are in very close to circular orbits and the circle is evenly
divisible by 30, 45, 60, and 120 if we divide the circle into 360 degrees which are special angles
very useful to the workings of physics and geometry. Further, the 360 degrees of a circle are
about the 365 days of a year (period of one earth orbit) so as such the earth moves through
about a degree a day in its journey around the sun. Thus, through these observations down
through the ages since ancient times we have constructed the duration of a second wisely
enough to make a lot work together. Now we see 6 protons, which is carbon the core element
of biological life on the planet where all of this came together is deeply connected with the
second that defines it all. With this idea of proton-seconds describing hydrogen and carbon the
basis of life the hydrocarbons in a cycle of 6 with respect to one another, the motion of the
earth around the sun and moon around the earth, and the basis of geometry the 360 degree
circle, equation 1 connects them with the universal constant of gravitation, the speed of light,
the fine structure constant and Planck’s constant that characterizes the physics of the atom. I
really wonder if other star systems are connected so well their planets with their star as the
Earth with the Sun. We really can’t resolve planets around other stars because the stars are too
far and so bright compared to their orbiting planets and the planets have to be large enough to
be inferred by the motion they induce in the star they orbit. It is hard to do so with earth sized
planets that might harbor life. However that may change in the near future.#
Since the earth day gets longer at a rate of one hour per 0.0067 million years then, the earth
will be (24)(365.25)/(360)=24.35 hours per day when it goes through 360 days per year. This will
be in 0.35/0.0066=52.2388 million years or, the year 522,408,21.!
Cr Mg F
O C Be
L i He H
=
24 12 19
8 6 4
3 2 1
24(6 ⋅ 1 − 4 ⋅ 2) − 12(8 ⋅ 1 − 4 ⋅ 3) + 19(8 ⋅ 2 − 6 ⋅ 3) = − 24(2) + 12(4) − 19(−2)
= − 48 + 48 + 38 = stro nt iu m (Sr)
−48 + 48 = 0
−48
i + 48
j = 48
2
+ 48
2
= 9.797958971
9.797958971 = 3.13016916 ≈ π = 3.141592654
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I made a program that looks for close to whole number solutions, I set it at decimal part equal
to 0.25. You cam choose how may values for t you want to try, and by what to increments
them. Here are the results for incrementing by 0.25 seconds then 0.05 seconds. Constant to all
of this is hydrogen and carbon. The smaller integer value of seconds gives carbon (6 protons at
1 second) and the largest value integer values of seconds give hydrogen (1 proton at six
seconds) and outside of that for the other integer values of protons you run are at t>0 and t<1.
Equation 1 really has some interesting properties. Here are two runs of the program( depart is
just me verifying that my boolean test was working right to sort out whole number solutions):#
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no
more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
By what value would you like to increment?: 0.05
How many values would you like to calculate for t in equation 1 (no
more than 100?): 100
40.1998 protons 0.150000 seconds 0.199837 decpart
30.1499 protons 0.200000 seconds 0.149879 decpart
24.1199 protons 0.250000 seconds 0.119904 decpart
20.0999 protons 0.300000 seconds 0.099918 decpart
17.2285 protons 0.350000 seconds 0.228500 decpart
15.0749 protons 0.400000 seconds 0.074938 decpart
12.0599 protons 0.500000 seconds 0.059950 decpart
10.0500 protons 0.600000 seconds 0.049958 decpart
8.0400 protons 0.750000 seconds 0.039966 decpart
7.0941 protons 0.850000 seconds 0.094088 decpart
6.0300 protons 1.000000 seconds 0.029975 decpart
5.2435 protons 1.150000 seconds 0.243457 decpart
5.0250 protons 1.200000 seconds 0.024980 decpart
4.1586 protons 1.450000 seconds 0.158605 decpart
4.0200 protons 1.500000 seconds 0.019985 decpart
3.1737 protons 1.899999 seconds 0.173673 decpart
3.0923 protons 1.949999 seconds 0.092296 decpart
3.0150 protons 1.999999 seconds 0.014989 decpart
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2.2333 protons 2.699999 seconds 0.233325 decpart
2.1927 protons 2.749999 seconds 0.192719 decpart
2.1536 protons 2.799999 seconds 0.153564 decpart
2.1158 protons 2.849998 seconds 0.115782 decpart
2.0793 protons 2.899998 seconds 0.079303 decpart
2.0441 protons 2.949998 seconds 0.044061 decpart
2.0100 protons 2.999998 seconds 0.009993 decpart
1.2433 protons 4.850000 seconds 0.243294 decpart
1.2306 protons 4.900001 seconds 0.230607 decpart
1.2182 protons 4.950001 seconds 0.218177 decpart
Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27,
h=6.62607E-34,G=6.67408E-11, c=299792459,protons[100],r=0.833E-15;
int main(int argc, const char * argv[]);
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in
equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/
(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{
printf("%.4f protons %f seconds %f decpart \n",
protons[i], t-increment, decpart);
}}}}
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The Author: Ian Beardsley
Copyright © November 16 2021
